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Ask the Wizard #240

I have heard that to finance the health care bill, a surcharge will be imposed on GROSS income above a certain point. This will have a big impact on high-level slot players, who accumulate hundreds of W2-G forms, like me. Do you have any insight?

Joe from Denver

Here is what the bill says:

In the case of a taxpayer other than a corporation, there is hereby imposed (in addition to any other tax imposed by this subtitle) a tax equal to 5.4 percent of so much of the modified adjusted gross income of the taxpayer as exceeds $1,000,000. -- Section 59C(a) page 337 H.R. 3962 (PDF — 3270 KB) or
The surcharge would be applied before the gambler could deduct any offsetting losses. I verified this with Marissa Chien, co-author of Tax Help for Gamblers. For high-level slot players, it is not difficult to rack up W2-G forms in the millions per year. Most of these players will still have a net loss on an annual basis. Past the million point in gross income, the player will pay a 5.4% tax on any win of $1,200 or more, even if there is a net loss for the year. This is just my opinion, but I think that isn’t fair. If we must tax gambling winnings (which they don’t in Canada), it should be on the net, not the gross winnings, on an annual basis. Should this become law, it will ruin high-level slot play in this country.

Marissa is on Twitter at @taxpro4gamblers, where she occasionally answers tax questions to followers.

I played 66 hands of Pick ’Em Poker in Missouri and never had a made winner (pair of nines or better) on the deal. What are the odds of that?

Dave from Overland Park, KS

For those unfamiliar with the rules, in Pick ’Em Poker the player is dealt two cards, plus the choice of one of two more. The game then gives the player two more cards to complete a five-card poker hand. The question at hand is what is the probability of having at least a pair of nines on the deal. Let’s call the players initial two cards that he must keep the "pocket," and the other two cards the "field." This could be accomplished the following ways:

  • Four of a kind: 13 combinations
  • High (9-A) three of a kind: 1,152 combinations
  • Low (2-8) three of a kind with the singleton in the field: 672 combinations
  • Two high pairs: 540 combinations
  • One high pair, one low pair, with at least one high card in the pocket: 1,260 combinations
  • High pair, with at least one in the pocket: 31,680 combinations
The sum of those combinations is 35,317. The total number of ways to pick 4 cards out of 52 is combin(52,4)=270,725. So the probability of getting a pair of nines or better on the deal is 31,680/270,725 = 13.05%. The probability of not getting a pair of nines or better is 100%-13.05%=86.95%. The probability of going 66 hands without a pair of nines or better is (1-(31,680)/270,725))66 = 0.00009848, or 1 in 10,155. That could have just been ordinary bad luck, and doesn’t rise to the level to make a convincing case of foul play. A bigger sample size is warranted to make a better case.

I’m a maths teacher, and I wanted to use your site as part of an investigation into the Mathematics of Gambling. However, I am concerned that exposure to a discussion of gambling systems will encourage, rather than discourage, gambling. Do you have any recommendations for problem gamblers? Or ways to avoid becoming a problem gambler? Sorry this question isn’t simple and Mathematical :) I want to combine a knowledge of the maths with an ethical/political understanding of the issue. Australia has a big Pokie (slot machine) gambling problem.

Abigail from Brisbane, QLD, Australia

My philosophy is the world would be a better place if there was unfettered access to truthful information. In that spirit, I would have no compunction to discuss the topic. If everybody knew the truth about how slot machines work, and how expensive they are to play, there would be a lot fewer players -- recreational and compulsive.

I’m aware of Australia’s love of pokies. When I was at a gambling conference in Sydney, I had the pleasure of listening to your Nick Xenophon chastising the audience for making such an addictive product. Personally, I favor mandating that machines be labeled with the return percentage that they are expected to pay.

Here in the U.S., we would say "math teacher," or "knowledge of math," by the way.

This happened to me this week, and am eagerly curious as to the statistic. Over two nights, I held Pocket Aces 3 times in total, and all 3 times I had them there was another player on the table of 10 players also with Pocket Aces. I have not been able to find the probability of this happening anywhere and I hope you can shed some light on this. On a full table of 10 players, what is the chance of this happening?

Rob T. from Hong Kong

The probability of a specific other player having pocket aces, given that you do, is (2/50)×(1/49) = 1 in 1,225. Given 9 other players, the probability is 9 times that, or 1 in 136. This might seem like an abuse of taking the sum of probabilities. However, it is okay if only one player can get the two aces. To answer your question, the probability that another player had pockets aces three out of the three times you had pockets aces is (9×(2/50)×(1/49))3 = 1 in 2,521,626.

The Orleans has a side bet in roulette that pays 8 to 1 on three reds (or blacks) in a row. It is on a double-zero wheel. Can you tell me the odds?

Haig from Englewood

The probability of winning is (18/38)3 = 10.63%. The house edge is 8×0.1063 - 1×0.8937 = 4.34%, which is less than the 5.26% on all the other bets (except the dreaded 0,00,1,2,3 combination at 7.89%).